Universal covering space of X x classifying space of \pi_1(X)

Question

I am trying to learn about classifying spaces for a Lie group $G$. The question I have is the following:

Suppose $X$ is a manifold and $G=\pi_1(X)$ is its fundamental group, is it true that $p:\tilde{X}\rightarrow X$ is a universal principal $G$-bundle?

I think it is not true in general, because I read somewhere in the book " Fiber Bundles" by Husemoller that a classifying principal bundle (P,B,G) must have the property that P is contractible.

Can anyone please make things clearer for me?

Answer 1

This is false in general as Lee Mosher illustrated, but nevertheless true for an interesting class of examples, namely aspherical manifolds. Covering maps induce isomorphisms on all higher homotopy groups, so if the manifold $M$ is aspherical, it's universal covering space $\tilde{M}$ is aspherical too and furthermore simply connected, hence all homotopy groups vanish and since all manifolds have the homotopy type of CW complexes, the universal $\tilde{M}$ is contractible by the Whitehead theorem. That implies that the universal covering is a universal $\pi_1(M)$-bundle.

A large class of aspherical example come from the theorem of Cartan-Hadamard, which proves the asphericality of all manifolds admitting a complete non-positively curved Riemmannian metric.

In this context, it is also worth mentioning the Borel conjecture which basically says, that closed aspherical manifolds are topologically completely determined by their fundamental group.

Answer 2

A counter-example always has a clarifying effect: $\mathbb{R} P^2$ is a manifold, $\mathbb{Z}/2\mathbb{Z}$ is its fundamental group, and $p : S^2 \to \mathbb{R} P^2$ is its universal cover. And $S^2$, of course, is not contractible.